Here for integers $m\geq 1$ I denote the Euler's totient function as $\varphi(m)$ and the sum of positive divisors $\sum_{d\mid m}d$, as $\sigma(m)$.
I know that it is possible to prove the following claim.
Claim. If $n$ is an odd perfect number (when we are on assumption of the hypothesis that there exists one of them) then satisifies
$$4\varphi \left(2^{n-1} \sigma(n)\right) ^2=2^{\sigma(n)}\varphi(n)^2.\tag{1}$$
In this post I would like to explore questions related to previous equation. First I am going to ask you about a computational exercise/experiment.
Question 1 (A computational exercise). Can you find an integer $m\geq 1$ satisfying $$4\varphi \left(2^{m-1} \sigma(m)\right) ^2=2^{\sigma(m)}\varphi(m)^2?$$ If you cann't find it add up to what segment of integers you ran your program, that I am going to accept your answer for this computational exercise. Many thanks.
I've tested that there aren't solutions of $(1)$ up to $10^5$ with a Pari/GP program. Can you improve it to a higher $10^N$, or well argue up to what limit $10^N$ cann't we find a solution?
Now we continue our study about previous equation $(1)$ with next remark.
Remark. Seems$^{\dagger}$ that there exist many positive integers $m\geq 1$ satisfying the congruence relation $$4\varphi \left(2^{m-1} \sigma(m)\right) ^2\equiv 0\text{ mod }2^{\sigma(m)}\varphi(m)^2.\tag{2}$$
I would like to prove that there exist infinitely many solutions of $(2)$.
Question 2. Please provide hints or details about what work can be done to prove that the congruence relation $$4\varphi \left(2^{m-1} \sigma(m)\right) ^2\equiv 0\text{ mod }2^{\sigma(m)}\varphi(m)^2$$ has infinitely many solutions over positive integers $m\geq 1$. Many thanks.
I think that a good strategy is to prove that the primes of the form $p\equiv 1\text{ mod }4$ satisfy the mentioned congruence relation, and conclude from here, since we know the Dirichlet's theorem for primes on arithmetic progressions that tell us, in particular, that there exist infinitely many primes of the form $p=4\lambda+1$. From my strategy, thus, it sufficies to show that there exist infinitely many prime numbers $p$ of the form $p\equiv 1\text{ mod }4$ satisfying
$$2^{2p}\varphi \left( \frac{p+1}{2}\right)^2 \equiv0\text{ mod }2^{p+1}\cdot(p-1)^2.$$ And thus it sufficies to prove that there exist infinitely many primes $p$ of the form $p\equiv 1\text{ mod }4$ satisfying that $p-1$ divides $2^{(p-1)/2}\varphi(\frac{p+1}{2})$. But I don't know how finish the exercise (I tried also powers of two instead of primes of the form $4\lambda+1$, but I do not know how to solve it).
Appendix:
$\dagger$As curiosity also seems that there exist many positive integers $\hat{m}\geq 1$ satisfying $$4\varphi \left(2^{\hat{m}-1} \sigma(\hat{m})\right) ^2\neq 0\text{ mod }2^{\sigma(\hat{m})}\varphi(\hat{m})^2.\tag{3}$$
And also as remark about this condition $(3)$ is that many elements of the sequence A228870 from the OEIS satisfy $(3)$ (thus that the remainder is $\neq 0$).